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\topic{Lecture 7 \\Multiple Integral\\ \scriptsize Double Integration (29 Sep 2009)}
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\section{Double Integrals}
Consider the function $f(x,y)$, which is given for all $x, y$ in a closed bounded region $R$ of the $xy$-plane. We devide the region $R$ by drawing the parallel lines to the $x$ and $y$ axes. 

Let $(x_k,y_k)$ be any point inside the $k$th rectangle element, and $\Delta A_k$ is the area of the $k$th rectangle, then we form the sum 
\begin{equation}
I_k = \sum_{k=1}^{n}f(x_k,y_k).\Delta A_k
\end{equation}
Now as we increase the number of divisions taking smaller and smaller elementary areas. In this fashion , as $n$ tends to infinity and the dimensions of each subdivision tend zero, in such case we obtain a sequence of real numbers $I_{n_1}, I_{n_2},...$. Assuming that $f(x,y)$ is continuous, or piecewise continous in $R$, and $R$ is bounded by finitely many smooth curves, then this sequence converges and its limit is independent of the choice of the subdivisions and corresponding points $(x_k,y_k)$. This limit is called the \textbf{double integral} of $f(x,y)$ over the region $R$, and is denoted by the symbol
\[
\int\int_Rf(x,y)dxdy
\]
Thus,
\begin{equation}
\label{eq:DIDef}
I=\int\int_Rf(x,y)dxdy = \lim_{n\longrightarrow \infty} \sum_{k=1}^{n}f(x_k,y_k).\Delta A_k
\end{equation}
\subsection{Evaluation of Double Integrals}
Double integrals over a region $R$ may be evaluated in two ways as follows.
\subsubsection{First Method}
Suppose $R$ is the region bounded by the curves $y_1=g(x)$, $y_2=h(x)$, $x_1=a$ and $x_2=b$, such that 
\[
a \leq x \leq  b, ~~~~y_1 \leq  y  \leq y_2
\]
then 
\[
\int\int_Rf(x,y)dxdy = \int_a^b \left[\int_{y_1}^{y_2} f(x,y)dy \right]dx
\]

We first integrate the inner integral 
\[
\int_{y_1}^{y_2} f(x,y)dy
\]
In this integration, we keep $x$ fixed, that is we regard $x$ as a constant. The result of this integration will be a function of $x$, which is further integrated from $a$ to $b$ to obtain the value of double integral.
\subsubsection{Second Method}
Similarly, if $R$ is the region bounded by the curves $x_1=\phi(y)$, $x_2= \psi(y)$, $y_1=c$ and $y_2=d$, such that 
\[c \leq y \leq  d, ~~~~x_1 \leq  x  \leq x_2\]
then 
\[
\int\int_Rf(x,y)dxdy = \int_c^d \left[\int_{x_1}^{x_2} f(x,y)dx \right]dy
\]

In this case, we first integrate the inner integral 
\[
\int_{x_1}^{x_2} f(x,y)dx
\]
In this integration, we keep $y$ fixed, that is we regard $y$ as a constant. The result of this integration will be a function of $y$, which is further integrated from $c$ to $d$ to obtain the value of double integral.

In some cases we find that $R$ can not be written by any of these two inequalities, then we may divide region in finitely many portions having such inequality form. Then we integrate $f(x,y)$ over each portion and finally add all results which produces the value of the double integral of $f(x,y)$ over that region $R$.
\begin{example}
Evaluate 
\[
\int_0^1 \int_0^{\sqrt{1+x^2}} \frac{dxdy}{1+x^2+y^2} 
\]
\end{example}
\[ I = \int_0^1 \int_0^{\sqrt{1+x^2}} \frac{dxdy}{1+x^2+y^2} = \int_0^1 \frac{1}{\sqrt{1+x^2}} \left[ \tan^{-1} \frac{y}{\sqrt{1+x^2}} \right]_0^{\sqrt{1+x^2}} dx\]

\[=\int_0^1 \frac{1}{\sqrt{1+x^2}} [\tan^{-1}1 -\tan^{-1}0 ]  dx = \frac{\pi}{4} \int_0^1 \frac{dx}{\sqrt{1+x^2}}\]
\[= \frac{\pi}{4} [\log {x+\sqrt{1+x^2}}]_0^1 = \frac{\pi}{4} [\log {1+\sqrt{2}}]\]

\begin{example}
Evaluate $\int \int xy dxdy$ over the region in the positive quadrant for which $x+y \leq 1$. 
\end{example}

The region of integration is the area $A$ bounded by the two axes and the straight line $x+y=1$. We can consider it as the region bounded by the lines $y=0$, $y=1-x$, $x=0$, and $x=1$.

Therefore  
\[  \int \int_A xy dxdy = \int_0^1 \int_0^{1-x} xy dxdy = \int_0^1 x \left[ \frac{y^2}{2} \right]_0^{1-x} dx\]
\[= \frac{1}{2} \int_0^1 x(1-x)^2 dx = \frac{1}{2} \int_0^1 (x+x^3-2x^2) dx = \frac{1}{2} \left[\frac{x^2}{2} - \frac{2x^3}{3}+ \frac{x^4}{4}\right]_0^1 = \frac{1}{24} \]

\begin{enumerate}
\item  Evaluate $\int \int (x+y)^2 dxdy$ over the area bounded by the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\]
\item  
Evaluate the following double integrals.
{\centering}
\subprob $ \int_0^1 \int_0^2 (x+2) dx dy$
\subprob $ \int_0^a \int_1^b \frac{dxdy}{xy}$
\subprob $ \int_1^2 \int_0^x \frac{dxdy}{x^2+y^2}$
\subprob $ \int_0^1 \int_x^{\sqrt{x}} (x^2+y^2)dxdy$
\subprob $ \int_0^1 \int_0^{x^2} exp(y/x) dxdy$
\item  Evaluate $\int \int x^2 y^2 dxdy$ over the region $x^2+y^2 \leq 1$
\item  Evaluate $\int _{0}^{1}dx \int _{0}^{x} e^{y/x} dy$ 
\item  Evaluate $\int _{0}^{1} \int _{0}^{\sqrt{1+x^2}} \frac{dydx}{1+x^2+y^2}$      
\item  Evaluate $\int \int _{R} xy  dxdy$, Where $R$ is bounded by the quadrant of circle $x^{2} + y^{2} = a^{2}$, $x \ge 0$ and $y \ge 0$.
\item  Evaluate $\int \int _{A} xy dx dy$, Where $A$ is the domain bounded by $x-$axis, ordinate $x =2a$ and the curve $x^{2}= 4ay$.
\item  Evaluate $\int \int _{S}\sqrt{xy-y^{2}}dydx$, Where $S$ is a triangle with vertices $(0,0)$, $(10,1)$ and $(1,1)$.
\item  Evaluate $\int \int _{A} x^{2} dxdy$, where $A$ is the region in the first quadrant bounded by the hyperbole $xy = 16$ and the lines $y = x$, $y = 0$ and $x = 8$.
\item  Evaluate$\int \int (x^{2} +y^{2})dxdy$ throughout the area enclosed by the curves $y = 4x$, $x+y=3$, $y= 0$ and $y=2$.
\item  Evaluate $\int _{0}^{\pi /2}\int _{a(1-\cos \theta )}^{a} r^2 dr d\theta $ 
\end{problems}
\end{enumerate}





